The two modes of the 8086/8088, minimum mode and maximum mode, have nothing to do with segment arrangement. The modes determine which pins on the chip have which functions. In minimum mode, certain pins are sacrificed in favor of a smaller implementation. In maximum mode, those pins take on other functions, and the original functions are generated by the Bus Controller chip.
8086 has memory divided into segments.Each segment has its particular register like ES, DS, SS, CD (extra segment, data segment, stack segment and code segment).These registers hold the base address BA.Now, there two base registers(BX and BP) and two index registers(SI and DI) in 8086. These registers hols the effective address EA.Now Physical address PA is sum of EA and BA.That is,PA = BA+EA
Write both modes
The 8088 is slower than the 8086 because the 8088 is running an 8-bit bus, while the 8086 runs a 16-bit bus. The two processors are the same, 16-bit processors, but the 8088 requires twice as many memory accesses to do the same amount of work as the 8086.
you add the modes that you have found together and then divide that by two
The 8088 microprocessor has a total of 14 registers. This includes eight general-purpose registers (AX, BX, CX, DX, SP, BP, SI, DI), four segment registers (CS, DS, SS, ES), and two pointer registers (IP and flags register). These registers serve various purposes, such as arithmetic operations, memory addressing, and control flow.
You find the number in between the two modes. It could be a decimal.
There can be two modes in a data set. For example, in the data set {0,1,2,3,3,4,5,5,9}, there are two modes: 3 and 5.
Yes, a set of data can have two modes. It is called bimodal.
The two modes are zoom mode and edit mode
If you have 3 modes, write down the three as your answer. If the mode has two numbers, just write one of the numbers down.Repeat on the other two modes.
If you're talking about measures of central tendency, if you have two modes put both of them down as the answer.
A segment, in the 8086/8088 is a 64kb chunk of memory addressable by any particular value in a segment register. Specifically, there are four segment registers, Code Segment, Data Segment, Stack Segment, and Extra Segment. Each is used in the context of a particular instruction by multiplying the segment register by 16 (left shift 4) and then adding the particular offset contained in the instruction.This gives access to 1Mb of memory (a 20 bit address bus) using only a 16 bit segment register and a 16 bit offset but, in only one instruction, you only have access to 64kb at a time. It would take two instructions to access any location in memory; one to load the segment register, and one to access the desired location.Note that, since the segment register is only left shifted by 4, that sequential segments overlap each other at a distance of 16 bytes. Note also that, in the 80386 and higher incarnations of the 8086/8088, that protected mode changed the meaning of a segment register, making it impossible to do simple 1Mb address computations unless you were in Virtual 386 mode or you were in flat 32 bit memory mode. (Almost all modern incarnations run in flat 32 bit mode or flat 64 bit mode, making the concept of segmented addressing obsolete.)